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  • HDU 6027 Easy Summation【简单相加||快速幂】

    Easy Summation

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
    Total Submission(s): 965    Accepted Submission(s): 389


    Problem Description
    You are encountered with a traditional problem concerning the sums of powers.
    Given two integers n and k. Let f(i)=ik, please evaluate the sum f(1)+f(2)+...+f(n). The problem is simple as it looks, apart from the value of n in this question is quite large.
    Can you figure the answer out? Since the answer may be too large, please output the answer modulo 109+7.
     

    Input
    The first line of the input contains an integer T(1T20), denoting the number of test cases.
    Each of the following T lines contains two integers n(1n10000) and k(0k5).
     

    Output
    For each test case, print a single line containing an integer modulo 109+7.
     

    Sample Input
    3 2 5 4 2 4 1
     

    Sample Output
    33 30 10
     

    Source

    注意:
    fast_pow(ll p, ll n)中p可能爆int需写成long long

    #include<iostream>
    #include<algorithm>
    #include<cmath>
    #include<cstdio>
    #include<cstdlib>
    #include<queue>
    #include<map>
    #include<set>
    #include<stack>
    #include<bitset>
    #include<numeric>
    #include<vector>
    #include<string>
    #include<iterator>
    #include<cstring>
    #include<ctime>
    #include<functional>
    #define INF 0x3f3f3f3f
    #define ms(a,b) memset(a,b,sizeof(a))
    #define pi 3.14159265358979
    #define mod 1000000007
    #define lson l,m,rt<<1
    #define rson m+1,r,rt<<1|1
    using namespace std;
    
    typedef pair<int, int> P;
    typedef long long ll;
    typedef unsigned long long ull;
    const int maxn = 55;
    
    
    ll fast_pow(ll p, ll n)
    {
    	ll ans = 1;
    	while (n)
    	{
    		if (n & 1) ans = (ans*p) % mod;
    		p = (p*p) % mod;
    		n >>= 1;
    	}
    	return ans;
    }
    
    int main()
    {
    	int t;
    	scanf("%d", &t);
    	while (t--)
    	{
    		ll n, k, as = 0;
    		scanf("%lld%lld", &n, &k);
    		for (int i = 1; i <= n; i++)
    		{
    			as = (as + fast_pow(i, k)) % mod;
    		}
    		printf("%lld
    ", as);
    	}
    }


    Fighting~
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  • 原文地址:https://www.cnblogs.com/Archger/p/12774772.html
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